Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}8x+y &= 1 \\ 8x-y &= -5\end{align*}$
Solution: Begin by moving the $y$ -term in the second equation to the right side of the equation. $8x = y-5$ Divide both sides by $8$ to isolate $x$ $x = {\dfrac{1}{8}y - \dfrac{5}{8}}$ Substitute this expression for $x$ in the first equation. $8({\dfrac{1}{8}y - \dfrac{5}{8}}) + y = 1$ $y - 5 + y = 1$ Simplify by combining terms, then solve for $y$ $2y - 5 = 1$ $2y = 6$ $y = 3$ Substitute $3$ for $y$ in the top equation. $8x+ 3 = 1$ $8x+3 = 1$ $8x = -2$ $x = -\dfrac{1}{4}$ The solution is $\enspace x = -\dfrac{1}{4}, \enspace y = 3$.